Marnix.VanDaele@UGentmvdaele/voordrachten/zaragoza... · 2007. 11. 19. · 0 6 −1 0 1 ν2 R 11...

P-stable exponentially fitted

Obrechkoff methodsfor y′′ = f (x, y)

Marnix Van Daele, G. Vanden [email protected]

Vakgroep Toegepaste Wiskunde en Informatica

Universiteit Gent

Zaragoza, November 2007 – p.1/60

OutlineIntroduction on exponentially fitted (EF) methods

2-step Obrechkoff methods fory′′ = f(x, y)

P-stable Obrechkoff methods fory′′ = f(x, y)

P-stable EF Obrechkoff methods fory′′ = f(x, y)

Conclusions


Exponentially fitted methodsSince about 1990, the research group of Guido Vanden Berghe at

Ghent University did a lot of work on EF methods.

interpolation :a cos ω x + b sinω x +n−2∑

i=0

ci xi

quadrature (Newton-Cotes, Gauss)

differentiation

integration (linear multistep methods, Runge-Kutta(-Nystrom)

methods for IVP and BVP, eigenvalue problems, ...)

integral equations (Fredholm, Volterra)

. . .

Aim : build methods which perform very good when the solution has a

known exponential of trigonometric behaviour.


Exponentially fitted methodsresearchers involved at Ghent University:

Guido Vanden Berghe, Hans De Meyer, Jan Vanthournout,

Marnix Van Daele, Philippe Bocher, Hans Vande Vyver,Veerle

LedouxandDavy Hollevoet

collaboration with L. Ixaru


Linear multistep methodsA well known method to solve

y′′ = f(y) y(a) = ya y′(a) = y′a

is theNumerov method(order 4)

yn+1 − 2 yn + yn−1 =h2

12(f(yn−1) + 10 f(yn) + f(yn+1))

Construction : imposeL[z(x);h] = 0 for z(x) = 1, x, x2, x3, x4

where

L[z(x);h] := z(x + h) + α0 z(x) + α−1 z(x − h)

−h2(

β1 z′′(x + h) + β0 z′′(x) + β−1 z′′(x − h))


Exponential fittingConsider the initial value problem

y′′ + ω2 y = g(y) y(a) = ya y(a) = y′a .

If |g(y)| ≪ |ω2 y| then

y(x) ≈ α cos(ω x + φ)

To mimic this oscillatory behaviour, one could replace polynomials by

trigonometric (in the complex case : exponential) functions.


EF Numerov methodConstruction : imposeL[z(x);h] = 0 for

z(x) = 1, x, x2, sin(ω x), cos(ω x)

L[z(x);h] := z(x + h) + α0 z(x) + α−1 z(x − h)

−h2(

β1 z′′(x + h) + β0 z′′(x) + β−1 z′′(x − h))

yn+1 − 2 yn + yn−1 = h2 (λf(yn−1) + (1 − 2 λ) f(yn) + λ f(yn+1))

λ =1

4 sin2 θ2

−1

θ2=

1

12+

1

240θ2 +

1

6048θ4 + . . . θ := ω h


EF methodsGeneralisation : to determine the coefficients of a method, we impose

conditions on a linear functional. These conditions are related to

polynomials :

xq|q = 0, . . . , K

exponential or trigonometric functions, multiplied with powers of

x :

xq exp(±µx)|q = 0, . . . , P

or, with ω = i µ,

xq cos(ωx), xq sin(ωx)|q = 0, . . . , P

Classical method :P = −1


Choice ofωlocal optimization : based on local truncation error (lte)

e.g. for the EF Numerov method

y(xn+1) − yn+1 = −h6

240

(

y(6)(xn) + ω2 y(4)(xn))

+ . . .

=⇒ ω2n = −

y(6)(xn)

y(4)(xn)

ω is step-dependent


Choice ofωglobal optimization

Preservation of geometric properties (periodicity, energy, . . . )

M. VAN DAELE, G. VANDEN BERGHE, GEOMETRIC

NUMERICAL INTEGRATION BY MEANS OF EXPONENTIALLY

FITTED METHODS, APNUM 57 (2007) 415-435

Several means to give a value toω :

backward error analysis

linearisation : rewritey′′ = f(y) asy′′ + ω2 y = g(y) with

|g(y)| small

Hamiltonian : minimize the leading term inHn+1 − Hn

ω is constant over the interval of integration


Obrechkoff methods

Nikola Obrechkoff (1896-1963)

Obrechkoff methods (OM) :1940 for quadrature

Milne : OM for solving diff. eq. :1949


Obrechkoff methods for y′′ = f(x, y)

Two-step methods

yn+1 − 2yn + yn−1 =

m∑

i=1

h2 i(

βi0 y(2i)n+1 + 2βi1 y(2i)

n + βi0 y(2i)n−1

)

L[z(x);h] := z(x + h) − 2 z(x) + z(x − h)

−m∑

i=1

h2 i(

βi0 z(2i)(x + h) + 2βi1 z(2i)(x) + βi0 z(2i)(x − h))

symmetric method :L[z(x);h] ≡ 0 if z(x) is odd

L[1;h] ≡ 0

orderp ⇐⇒ L[xq;h] = 0, q = 0, 1, . . . , p + 1

lte = Cp+2 hp+2y(p+2)(xn) + O(

hp+3)

Cp+2 =L[xp+2;h]

(p + 2)!hp+2


Two-step OM

m = 1 : p = 4, C6 = −1

240(Numerov method)

yn+1 − 2yn + yn−1 =

h2

12

(

y(2)n+1 + 10 y(2)

n + y(2)n−1

)

m = 2 : p = 8, C10 =59

76204800

yn+1 − 2yn + yn−1 =

h2

252

(

11 y(2)n+1 + 230 y(2)

n + 11 y(2)n−1

)

−h4

15120

(

13 y(4)n+1 − 626 y(4)

n + 13 y(4)n−1

)


Two-step OM

m = 3 : p = 12, C14 = −45469

1697361329664000

yn+1 − 2yn + yn−1 =

h2

7788

(

229 y(2)n+1 + 7330 y(2)

n + 229 y(2)n−1

)

−h4

25960

(

11 y(4)n+1 − 1422 y(4)

n + 11 y(4)n−1

)

+h6

39251520

(

127 y(6)n+1 + 4846 y(6)

n + 127 y(6)n−1

)

. . .

method of orderp = 4m


EF 2-step OM : the casem = 2

yn+1 − 2yn + yn−1 =

h2(

β10 y(2)n+1 + 2β11 y(2)

n + β10 y(2)n−1

)

+h4(

β20 y(4)n+1 + 2β21 y(4)

n + β20 y(4)n−1

)

5 possibilities:

P = −1 : x2, x4, x6, x8

P = 0 : x2, x4, x6, cos ω x

P = 1 : x2, x4, cos ω x, x sin ω x

P = 2 : x2, cos ω x, x sinω x, x2 cos ω x

P = 3 : cos ω x, x sin ω x, x2 cos ω x, x3 sin ω x


EF 2-step OM : the casem = 2

yn+1 − 2yn + yn−1 =

h2(

β10 y(2)n+1 + 2β11 y(2)

n + β10 y(2)n−1

)

+h4(

β20 y(4)n+1 + 2β21 y(4)

n + β20 y(4)n−1

)

5 possibilities:

P = −1 : x2, x4, x6, x8

β21 =313

15120β20 = −

13

15120β11 =

115

252β10 =

11

252


EF 2-step OM : the casem = 2P = 0 : x2, x4, x6, cos ω x

β10 =1

30−12 β20 β21 =

1

40+5β20 β11 =

7

15+12β20

β20 = −120 − 4 cos (θ) θ2 − 56 θ2 − 120 cos (θ) + 3 θ4

120 θ2 (12 cos (θ) − 12 + cos (θ) θ2 + 5 θ2)

10 20 30−0.007

0

θ

β 20

continuous coefficients


EF 2-step OM : the casem = 2P = 1 : x2, x4, cos ω x, x sin ω x

β10 =1

12− 2 β20 − 2 β21 β11 =

5

12+ 2β20 + 2β21

β20 =θ5 sin θ + 2 (cos θ + 5) θ4 + 48 (cos θ − 1) θ2 + 48 (cos θ − 1)2

12 θ4 (θ3 sin θ − 4 (1 − cos θ)2)

β21 =5 θ5 sin θ − 2 cos θ (cos θ + 5) θ4

− 48 cos θ (cos θ − 1) θ2− 48 (cos θ − 1)2

12 θ4 (θ3 sin θ − 4 (1 − cos θ)2)

10 20 30−0.03

0

0.03

θ

β 20

discontinuous coefficients


EF 2-step OM : the casem = 2P = 2 : x2, cos ω x, x sinω x, x2 cos ω x

discontinuous coefficients

P = 3 : cos ω x, x sin ω x, x2 cos ω x, x3 sin ω x

continuous coefficients


Stability of two-step OMyn+1 − 2 yn + yn−1 =

m∑

i=1

h2 i(

βi0 y(2i)n+1 + 2βi1 y(2i)

n + βi0 y(2i)n−1

)

applied toy′′ = −λ2 y gives

yn+1 − 2Rmm(ν2) yn + yn−1 = 0 ν := λh

Rmm(ν2) =

1 +

m∑

i=1

(−1)iβi1 ν2i

1 +m∑

i=1

(−1)i+1βi0 ν2i

The stabilitity functionRmm uniquely determines the method.

A method has theinterval of periodicity(0, ν20) if

|Rmm(ν2)|≤1 for 0 < ν2 < ν20 .

The method isP -stableif |Rmm(ν2)|≤1 for all realν 6= 0.


Stability of two-step OM

m = 1 : p = 4, C6 = −1

240(Numerov) ν2

0 = 6

yn+1 − 2yn + yn−1 =

h2

12

(

y(2)n+1 + 10 y(2)

n + y(2)n−1

)

0 6−1

0

1

ν2

R11



m = 2 : p = 8, C10 =59

76204800ν20 = 25.2

yn+1 − 2yn + yn−1 =

h2

252

(

11 y(2)n+1 + 115 y(2)

n + 11 y(2)n−1

)

−h4

15120

(

13 y(4)n+1 − 626 y(4)

n + 13 y(4)n+1

)

0 25.2−1

0

1

ν2

R22



m = 3 : p = 12, C14 = −45469

1697361329664000ν20 = 55.60 . . .

yn+1 − 2yn + yn−1 =

h2

7788

(

229 y(2)n+1 + 7330 y(2)

n + 229 y(2)n−1

)

+h4

25960

(

−11 y(4)n+1 + 1422 y(4)

n − 11 y(4)n−1

)

+h6

39251520

(

127 y(6)n+1 + 4846 y(6)

n + 127 y(6)n−1

)

0 55.6062−1

0

1

ν2

R33


P-stable 2-step OMAnanthakrishnaiah (1987)

idea : use some parameters to obtain P-stability.

m = 3 andm = 4

e.g.m = 3

yn+1 − 2yn + yn−1 =

h2(

β10 y(2)n+1 + 2β11 y(2)

n + β10 y(2)n−1

)

+h4(

β20 y(4)n+1 + 2β21 y(4)

n + β20 y(4)n−1

)

+h6(

β30 y(6)n+1 + 2β31 y(6)

n + β30 y(6)n−1

)

imposep = 6 : x2, x4, x6,

then 3 parameters (β20, β30 andβ31) remain

R33 =1 − β11 ν2 + β21 ν4 − β31 ν6

1 + β10 ν2 − β20 ν4 + β30 ν6let β31 = β30


Ananthakrishnaiah’s ideaChoose these 2 remaining parametersβ20, β30 such that the method

becomes P-stable

|R33| =

∣

∣

∣

∣

N3

D3

∣

∣

∣

∣

< 1 ⇐⇒ (D3 − N3) (D3 + N3) > 0

0 0.001−0.008

0

0.004

β30

β 20


Ananthakrishnaiah’s ideaFind the solution for which the phase-lag,

ν − arccos R33 =

(

13

604800+

13

30β30 +

1

40β20

)

ν7 + O(

ν9)

becomes minimal.

0 0.001−0.008

0

0.004

β30

β 20


Ananthakrishnaiah’s ideaFind the solution for which the phase-lag,

ν − arccos R33 =

(

13

604800+

13

30β30 +

1

40β20

)

ν7 + O(

ν9)

becomes minimal.

0 0.001−0.008

0

0.004

β30

β 20

This gives(β30, β20) =

(

1

14400,−

1

600

)

.


Ananthakrishnaiah’s ideaAnanthakrishnaiahm = 3 : p = 6, P-stable,C8 = −

1

50400

yn+1 − 2yn + yn−1 =

h2

20

(

y(2)n+1 + 18 y(2)

n + y(2)n−1

)

−h4

600

(

y(4)n+1 − 22 y(4)

n + y(4)n−1

)

+h6

14400

(

y(6)n+1 + 2 y(6)

n + y(6)n−1

)


Ananthakrishnaiah’s idea

R33 =1 − 9

20 ν2 + 11600 ν4 − 1

14400 ν6

1 + 120 ν2 + 1

600 ν4 + 114400 ν6

0 20 40 60 80 100−1

0

1

ν

R

|R33| =

∣

∣

∣

∣

N3

D3

∣

∣

∣

∣

< 1 since

(D3 − N3) (D3 + N3) =ν2 (ν2 − 10)2 (ν2 − 60)2

360000


P-stable 2-step OMWe were able to generalise Ananthakrishnaiah’s idea in

M. VAN DAELE AND G. VANDEN BERGHE,

P-STABLE OBRECHKOFF METHODS OF ARBITRARY ORDER

FOR SECOND-ORDER DIFFERENTIAL EQUATIONS, NUMERICAL ALGORITHMS 44, 2007,

115-131

Algoritm to construct a P-stable OM for a givenm :

impose order2 m

Dm − Nm andDm + Nm should both be halves of perfect

squares

This leads to a system ofnon-linearequations.

Theorem : the approximantRmm(ν2), obtained by generalising

Ananthakrishnaiah’s approach, is given by the real part of the

(m, m)-Padé approximant ofexp(i ν).

This leads to a system oflinearequations.Zaragoza, November 2007 – p.30/60

Example

R33(ν2) =

1 − 920 ν2 + 11

600 ν4 − 114400 ν6

1 + 120 ν2 + 1

600 ν4 + 114400 ν6

= ℜ

(

1 + 12 i ν − 1

10 ν2 − 1120 i ν3

1 − 12 i ν − 1

10 ν2 + 1120 i ν3

)

=

(

1 − 110 ν2

)2−(

12 ν − 1

120 ν3)2

(

1 − 110 ν2

)2+(

12 ν − 1

120 ν3)2

where1 + 1

2 ν + 110 ν2 + 1

120 ν3

1 − 12 ν + 1

10 ν2 − 1120 ν3

= exp(ν) + O(

ν7)

|R33(ν2)| =

∣

∣

∣

∣

∣

(

1 − 110 ν2

)2−(

12 ν − 1

120 ν3)2

(

1 − 110 ν2

)2+(

12 ν − 1

120 ν3)2

∣

∣

∣

∣

∣

≤ 1


ConclusionHow does the P-stable Obrechkoff method

yn+1 − 2yn + yn−1 =m∑

i=1

h2 i(

βi0 y(2i)n+1 + 2βi1 y(2i)

n + βi0 y(2i)n−1

)

of order2 m look like ?

βi0 = (−1)i+1a2i + 2

i−1∑

j=0

(−1)j+1aj a2i−j

βi1 = a2i + 2

i−1∑

j=0

aj a2i−j

i = 1 . . . , m

whereaj =

(m

j )(2 m

j )for 0 ≤ j ≤ m

0 for j > m


P-stable Expon. fitted OMHow to obtain P-stable exponentially-fitted Obrechkoff methods ?

yn+1 − 2 yn + yn−1 =m∑

i=1

h2 i(

βi0 y(2i)n+1 + 2βi1 y(2i)

n + βi0 y(2i)n−1

)

applied toy′′ = −λ2 y gives

yn+1 − 2Rmm(θ, ν2) yn + yn−1 = 0

with θ := ω h andν := λh

Padé-approximants=⇒ exponentially-fitted Padé approximants

Construction of Polynomially-fitted(m, m) Padé approximants :

Pm(x)

Pm(−x)= exp(x) + O

(

x2 m+1)

⇐⇒ exp(x) Pm(−x) − Pm(x) = O(

x2 m+1)

⇐⇒d2 q

dx2 q (exp(x) Pm(−x) − Pm(x))∣

∣

∣

x=0= 0 q = 1, . . . , m


EF Padé approximants

F(x, t) = exp(t x) Vm(−t x)−Vm(t x) Vm(x) = 1+

m∑

j=1

aj xj

∂2q

∂x2q F(x, t)∣

∣

∣

(x,t)=(0,θ)= 0 q = 1, . . . , K

ℜ

(

∂q

∂tqF(x, t)

∣

∣

∣

(x,t)=(i, θ)

)

= 0 q = 0, . . . , P

where0 ≤ K ≤ m andP + K + 1 = m.

This leads to a system ofm linear equations in the unknownsaj ,

j = 1, . . . , m.

The EF(K, P ) Padé approximant toexp(ν) is then given by(K,P )Pm

m (ν) = Vm(ν)/Vm(−ν).


EF Padé approximants :m = 1

F(x, t) = exp(t x) V1(−t x) − V1(t x) Vm(x) = 1 + a1 x

(K = 1, P = −1)

∂2

∂x2 F(x, t)∣

∣

∣

(x,t)=(0,θ)= 0 ⇐⇒ θ2 (1 − 2 a1) = 0 ⇐⇒ a1 =

1

2

(1,−1)P 11 (ν) =

1 + 12 x

1 − 12 x(K = 0, P = 0)

ℜ (F(i, θ)) = 0 ⇐⇒ ℜ(

ei θ (1 − i a1 θ) − (1 + i a1θ))

= 0

⇐⇒ a1 =sin θ

θ (cos θ + 1)=

1

2

tan(θ/2)

θ/2

(0,0)P 11 (ν) =

1 + 12

tan(θ/2)θ/2 x

1 − 12

tan(θ/2)θ/2 x



m = 1

(K, P ) a1

(1,−1)1

2

(0, 0)tan θ

2

θ



m = 1

(K, P ) a1

(1,−1)1

2

(0, 0)1

2+

θ2

24+ O(θ4)



m = 1

(K, P ) a1

(1,−1)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

(0,0)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1



m = 2

(K, P ) a1 a2

(2,−1)1

2

1

12

(1, 0)1

2

2 tan θ

2− θ

2 θ2 tan θ

2

(0, 1)2 (1 − cos θ)

(θ + sin θ) θ

θ − sin θ

(θ + sin θ) θ2



m = 2

(K, P ) a1 a2

(2,−1)1

2

1

12

(1, 0)1

2

1

12+

θ2

720+ O

(

θ4)

(0, 1)1

2+ O(θ4)

1

12+

θ2

360+ O(θ4)



m = 2

(K, P ) a1 a2

(2,−1)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

0 2π 4π 6π 8π 10π

−0.1

−0.05

0

0.05

0.1

θ

a 2

(1, 0)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 2

(0, 1)

0 2π 4π 6π 8π 10π0

0.1

0.2

0.3

0.4

0.5

θ

a 1

0 2π 4π 6π 8π 10π0

0.02

0.04

0.06

0.08

0.1

θ

a 2


m = 3

(K, P ) a1 a2 a3

(3,−1)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

0 2π 4π 6π 8π 10π

−0.1

−0.05

0

0.05

0.1

θ

a 2

0 2π 4π 6π 8π 10π−0.015

−0.01

−0.005

0

0.005

0.01

θ

a 3

(2, 0)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

0 2π 4π 6π 8π 10π

−0.1

−0.05

0

0.05

0.1

0.15

θ

a 2

0 2π 4π 6π 8π 10π−0.05

0

0.05

θ

a 3

(1, 1)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

0 2π 4π 6π 8π 10π0

0.02

0.04

0.06

0.08

0.1

θ

a 2

0 2π 4π 6π 8π 10π0

0.005

0.01

0.015

θ

a 3

(0, 2)

0 2π 4π 6π 8π 10π−1

−0.5

0

0.5

1

θ

a 1

0 2π 4π 6π 8π 10π

−0.1

−0.05

0

0.05

0.1

0.15

θ

a 2

0 2π 4π 6π 8π 10π−0.05

0

0.05

θ

a 3



−1 −0.5 0 0.5 10

0.5

1

1.5

2

2.5

ν

exp(ν)(1,−1)P

11(ν)

(0,0)P11(ν), θ=1

(0,0)P11(ν), θ=2

(0,0)P11(ν), θ=4



−1 0 1 2 3 40

10

20

30

40

50

ν

θ=0

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)



−1 0 1 2 3 40

10

20

30

40

50θ=1

ν

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)



−1 0 1 2 3 40

10

20

30

40

50θ=2

ν

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)



−1 0 1 2 3 40

10

20

30

40

50θ=4

ν

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)



−1 0 1 2 3 40

10

20

30

40

50θ=7

ν

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)



−1 −0.5 0 0.5 10

0.5

1

1.5

2θ=7

ν

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)

−1 0 1 2 3 40

10

20

30

40

50θ=7

ν

exp(ν)(3,−1)P

33(ν)

(2,0)P33(ν)

(1,1)P33(ν)

(0,2)P33(ν)


The Stiefel-Bettis problem

z′′ + z = 0.001 exp(ix) z(0) = 1 z′(0) = 0.995i 0 ≤ x ≤ 40π

equivalent real form

u′′ + u = 0.001 cos(x), u(0) = 1, u′(0) = 0,

v′′ + v = 0.001 sin(x), v(0) = 0, v′(0) = 0.9995 .

z(x) = u(x) + i v(x)

u(x) = cos(x) + 0.0005x sin(x)

v(x) = sin(x) − 0.0005x cos(x)

γ(x) =√

u2(x) + v2(x) =√

1 + (0.0005x)2

E := maxxj∈[0, 40 π]

‖y(xj) − yj‖ h = 2−i π ω = 1Zaragoza, November 2007 – p.50/60

The Stiefel-Bettis problemm = 1

−1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0−15

−10

−5

0

log10(h)

log1

0(E

)

: P = −1 ⋄ : P = 0Zaragoza, November 2007 – p.51/60


−1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0−16

−14

−12

−10

−8

−6

−4

−2

0

log10(h)

log1

0(E

)

: P = −1 ⋄ : P = 0 : P = 1Zaragoza, November 2007 – p.52/60


−1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0−15

−10

−5

0

log10(h)

log1

0(E

)

: P = −1 ⋄ : P = 0 : P = 1 ⋆ : P = 2Zaragoza, November 2007 – p.53/60

The Duffing problem

y′′(x) = −y(x) − y(x)3 + cos(Ωx) 0 ≤ x ≤ 50π

B = 0.002 Ω = 1.01

y(x) =

4∑

i=0

K2i+1 cos[(2i + 1)Ωx] ,

(K1, K3, K5, K7, K9) =(

0.20017947753661502, 2.46946143255559 10−4,

3.0401498519692437 10−7, 3.743490701609247 10−10,

4.609682949622697 10−13)


The Duffing problemy(3)(x) = −(1 + 3 y2(x)) y′(x) − B Ω sin(Ωx)

y(4)(x) = −(1 + 3 y2(x)) y′′(x) − 6y(x)y′(x)2 − B Ω2 cos(Ωx)

y(5)(x) = −(1 + 3 y2(x)) y′′′(x) − 6y′(x)3 − 18y(x)y′(x)y′′(x)

+B Ω3 sin(Ωx)

y(6)(x) = −(1 + 3 y2(x)) y(4)(x) − 24y(x)y′(x)y(3)(x) − 18y(x)y′′(x)2

−36y′(x)2y′′(x) + B cos(Ω x)Ω4

y(7)(x) = −(1 + 3 y2(x)) y(5)(x) − 30 y(x) y(4)(x)

−60 (y′(x) + y(x) y′′(x)) y′′′(x) − B sin(Ω x)Ω5− 90 y′(x) y′′(x)2

z′′ =

y

y′

′′

=

f(x, y)

g(x, y, y′)

g(x, y, y′) =ddx

f(x, y(x))

E := maxxj∈[0, 50 π] ‖z(xj) − zj‖

h = π/2i, i = 2, 3, . . . , 7 ω = Ω


The Duffing problemm = 1

−2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4−15

−10

−5

0

log10(h)

log1

0(E

)

: P = −1 ⋄ : P = 0Zaragoza, November 2007 – p.56/60


−2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4−15

−10

−5

0

log10(h)

log1

0(E

)

: P = −1 ⋄ : P = 0 : P = 1Zaragoza, November 2007 – p.57/60


−2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4−15

−10

−5

0

log10(h)

log1

0(E

)

: P = −1 ⋄ : P = 0 : P = 1 ⋆ : P = 2Zaragoza, November 2007 – p.58/60

The Duffing problemm = 3, (K, P ) = (2, 0)

−2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4−15

−10

−5

0

log10(h)

log1

0(E

)

ω = 0 (circles) ω = 1.01 (diamonds) ω = 0.7 (squares)

ω = 2 (asterisks)Zaragoza, November 2007 – p.59/60

ConclusionTwo-step P-stable exponentially fitted Obrechkoff methods :

for any givenm, P-stable EF(K, P )-methods of order2 m exist

the construction is based on EF Padé approximants to the

exponential function

the coefficients depend on a parameterθ

the coefficients are continous functions ofθ iff P is odd

numerical examples are given as an illustration

References :

M. Van Daele and G. Vanden Berghe, P-stable Obrechkoff methods of arbitrary order for

second-order differential equations, Numerical Algorithms44, 2007, 115-131

M. Van Daele and G. Vanden Berghe, P-stable exponentially fitted Obrechkoff methods

of arbitrary order for second-order differential equations, accepted for publication in

Numerical Algorithms


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